RESOLVENT POSITIVE OPERATORS
WOLFGANG ARENDT
[Received 15 October 1985—Revised 14 February 1986]
ABSTRACT
Resolvent positive operators on an ordered Banach space (with generating and normal positive
cone) are by definition linear (possibly unbounded) operators whose resolvent exists and is positive on
a right half-line. Even though these operators are defined by a simple (purely algebraic) condition,
analogues of the basic results of the theory of Q-semigroups can be proved for them. In fact, if A is
resolvent positive and has a dense domain, then the Cauchy problem associated with A has a unique
solution for every initial value in the domain of A2, and the solution is positive if the initial value is
positive. Also the converse is true (if we assume that A has a non-empty resolvent set and D(A2) n E+
is dense in £+). Moreover, every positive resolvent is a Laplace-Stieltjes transform of a so-called
integrated semigroup; and conversely every such (increasing, non-degenerate) integrated semigroup
defines a unique resolvent positive operator.
1. Introduction
Many problems in applied mathematics occur in the form of a Cauchy problem
(
}
"(0)=/,
where A is a linear operator and / (the 'initial value') an element of its domain.
There is a well established notion of well-posedness of (CP) which is equivalent to
A being the generator of a Q-semigroup (see Goldstein [13, II. 1.2]).
The purpose of the present paper is to show that for Cauchy problems
involving an operator A on an ordered Banach space such that solutions with
positive initial values remain positive, a different notion of well-posedness and of
a generator seems to be more natural. In fact, we suggest considering the
following class of operators:
DEFINITION. Let E be an ordered Banach space whose positive cone is
generating and normal. An operator A on E is called resolvent positive if there
exists weR such that (w,«>)c:p(A) (the resolvent set of A) and R(X, A): =
1
This notion is purely algebraic and no norm condition is required. By the
Hille-Yosida theorem a densely defined resolvent positive operator A is the
generator of a (necessarily positive) Q-semigroup (if and) only if in addition
sup{||(A -a)nR(k, A)"\\: k>a,ne N} <°° for some a 2= w.
Even though resolvent positive operators do not generate a one-parameter
semigroup in general, they admit a satisfactory theory which parallels the theory
of semigroups to a high extent. We want to explain this in more detail.
The principal objects in the theory of one-parameter semigroups interact in the
following way. To every Q-semigroup one associates its generator whose
resolvent is given by the Laplace transform of the semigroup. Moreover,
A.M.S. (1980) subject classification: 47D05.
Proc. London Math. Soc. (3) 54 (1987) 321-649.
322
WOLFGANG ARENDT
generators of (^-semigroups are (essentially) characterized by the fact that the
associated Cauchy problem admits a unique solution for all initial values in the
domain of the operator (well-posedness).
In complete analogy we find that every positive resolvent is the LaplaceStieltjes transform of a unique so-called 'integrated semigroup', and conversely,
to every (non-degenerate, increasing) integrated semigroup we associate a unique
resolvent positive operator whose resolvent is the Laplace-Stieltjes transform of
this integrated semigroup. Concerning the Cauchy problem, the following holds
(with some reservation regarding a technical detail): a densely defined operator is
resolvent positive if and only if the associated Cauchy problem admits a unique
solution for every initial value / in the domain of A2 and the solution is positive if
/ i s positive.
The paper is organized as follows. In § 2, we discuss order conditions on the
space or the domain which imply that a resolvent positive operator with a dense
domain is 'automatically' the generator of a Q-semigroup. These are exceptional
phenomena. In fact, in §3 several natural examples of resolvent positive
operators which are not generators of a semigroup are constructed.
In the general theory which follows in §§4-8 the representation of positive
resolvents as Laplace—Stieltjes transforms is essential. It is proved by two
different approaches. One is based on the Hille-Yosida theorem and can be
applied when A has a dense domain (§4). The other depends on a vector-valued
version of Bernstein's theorem which we prove in § 5. Here we have to restrict
the space (allowing reflexive spaces, ZZ-spaces and c0), but it is no longer
necessary to assume that the domain of the operator is dense.
As we pointed out above, in our theory the semigroup is replaced by the
so-called integrated semigroup. The relations between this integrated semigroup
and the given resolvent positive operator are similar to those between a
semigroup and its generator. They are investigated in § 6. The homogeneous
Cauchy problem is considered in § 7, the inhomogeneous in § 8.
Finally, we characterize resolvent positive operators on a Banach lattice by
means of Kato's inequality (§9). In some aspects, this result is similar to the
Lumer-Phillips theorem for generators of Q-semigroups.
General assumptions. Throughout the paper E denotes a real ordered Banach
space whose positive cone E+ is generating and normal (that is, E = E+ — E+ and
E' = E+ — E'+, where E'+ denotes the dual cone). For example, E may be a
Banach lattice or the hermitian part of a C*-algebra. Moreover, we assume that
the norm on E is chosen in such a way that
(1.1)
±f*8
implies ||/|| * \\g\\ (f,geE)
(which can always be done). We also note that there exists a constant k > 0 such
that for all linear operators 5, T on E one has
(1.2)
0 ^ 5 ^ T implies ||S|| «s* ||r||.
We refer to [4] and [21] for more details on ordered Banach spaces.
Acknowledgement. I would like to thank Paul R. Chernoff for several
stimulating discussions. My thanks also go to Frank Neubrander and Hermann
Kellermann for many discussions on 'integrated semigroups', and to Hans Engler
for a valuable estimate.
RESOLVENT POSITIVE OPERATORS
323
2. Resolvent positive operators which are automatically
generators of a Co-semigroup
Let A be a resolvent positive operator. We introduce the following notation:
s(A) = inf{w eU: (w, oo) c p(A) and R(X, A)^0 for all A > w},
and D(A')+ = E'+nD(A')
(where in the second definition we assume A to be densely denned; then A'
denotes the adjoint of A). Let s(A) <X<\i. Then
R(X, A) - R([t, A) = (fi- A)/?(A, A)R(fi, A) s* 0.
Thus the function /?(•, A) is decreasing on (s(A), °°).
LEMMA
(2.1)
2.1. Lef A be a resolvent positive operator such that s(A) < 0. Then
/?(0, A) = R(k, A) + A/?(A, A)2 + A2fl(A, A) 3 +... + A""1i?(A, A)n
+ knR(X, A)nR(0, A)
for all n e N, A 2= 0. Consequently,
(2.2)
sup{\\XnR(X, A)nR(0, A)\\: neN,A^0}<«>.
Proo/. By the resolvent equation, R(0, A) = R(X, A) +kR(k, A)R(0, A)
(A >0). This is (2.1) for n = 1. Iterating this equation yields (2.1) for all n e N.
A subset C of £ + is called cofinal in £ + if for every f eE+ there exists g 6 C
such that/=£g. If (T(t))tss0 is a positive Q-semigroup with generator B, then the
type (or growth bound) (o(B) is defined by <w(5) = inf{w e R: there exists M 3s 1
such that ||7(r)|| ^Me"" for all f 2=0}. One always has s(B) *s <w(fl) < oo, but it
can happen that s(B)i^co(B) even if B generates a positive Q-group [14,26].
THEOREM 2.2. Let A be a densely defined resolvent positive operator. If D(A)+
is cofinal in E+ or if D(A')+ is cofinal in E'+, then A is the generator of a positive
QfSemigroup. Moreover, s(A) = co(A).
Proof (a) Assume that s(A)<0. We claim that A generates a bounded
Q-semigroup, if one of the conditions in the theorem is satisfied. We first assume
that D(A) is cofinal. L e t / e E+. Then there exists g e D(A)+ such that/ssg. Let
h = -Ag and k e E+ such that h =s k. Then
It follows from (2.1) that
A"/?(A, A)nf s= A"/?(A, A)nR(Q, A)k ^ R(0, A)k.
Hence
Since E = E+- E+, it follows that
{XnR(X,A)n:
324
WOLFGANG ARENDT
is strongly bounded; thus it is norm-bounded by the uniform boundedness
principle. The Hille-Yosida theorem implies that A generates a bounded
(^-semigroup.
If D(A')+ is cofinal in E+, consider f e E+, <t>eE'+. Then there exists ip e E'+
such that 0 ^R(0, A)'ip. Hence by (2.1),
, A)nf, R(Q, A)'y)
Since E+ and E'+ are generating, this implies that {KnR(X,A)n: A^O, neN) is
weakly bounded, and so norm-bounded. Again the Hille-Yosida theorem implies
the claim.
(b) If s(A) is arbitrary, consider B =A - w for some w >s(A). Then s(B) < 0,
and so by (a), B is the generator of a bounded semigroup (T(t))t90. Hence A
generates the semigroup (ewtT(t))t^0. Moreover a)(A) ^ w.
COROLLARY 2.3. Assume that intE+^0.
If A is a densely defined resolvent
positive operator, then A is the generator of a positive Co-semigroup and
s(A) =
Proof. Since inlE+^0
and D(A) is dense, there exists u eint E+C\D(A).
The set {u} is clearly cofinal in E+.
COROLLARY 2.4. Let A be a densely defined resolvent positive operator on
L1(X, fi) (where (X, pi) is a o-finite measure space). If there exists <j> e D(A') D
L (X, ix) such that (f>(x) ^e>0 for almost all xeX, then A is the generator of a
positive Co-semigroup.
REMARK. Corollary 2.3 has been proved in [2] and Corollary 2.4 by Batty and
Robinson [4] with a different approach using half-norms.
THEOREM 2.5. Let A be a densely defined resolvent positive operator. If there
exist Ao > s(A) and c > 0 such that
(2-4)
||*(Ao,i4)/||*c Il/H
(feE+),
then A is the generator of a positive Co-semigroup and s(A) = <o(A).
Proof. Let s(A) < w ^ Ao. Let B = A - w. Then s(B) < 0. Since R(Q, B) =
R(w, A)^R(X0) A), it follows from (2.4) and (1.2) that
for all f eE+. In particular,
B))ng\\
for all g e E+, A 5= 0, n e N. Since E = E+ - E+, it follows from (2.2) that the set
{XnR{X, B)n: n eN, A5=0} is strongly bounded and so norm-bounded. Thus by
the Hille-Yosida theorem, B—A — w generates a bounded positive Qsemigroup. Hence A is the generator of a positive (^-semigroup as well and
co(A) =£ w.
RESOLVENT POSITIVE OPERATORS
325
REMARK. Theorem 2.5 (except the assertion concerning the spectral bound) is
due to Batty and Robinson [4] (with a different proof).
THEOREM 2.6. Suppose that the norm is additive on the positive cone, that is,
11/ + *ll = 11/11 + \\g\\ for all f,geE+ (for example, E = L\X, **)). Let A be a
densely defined operator. Then the following assertions are equivalent:
(i) A generates a positive Co-group;
(ii) A and -A are resolvent positive and there exist A >max{5(y4), s(-i4)} and
c > 0 such that
(2.5)
||*(A,±i4)/||*c Il/H
JbraUfeE+.
Proof Assume that A generates a positive Q-group (T(t))teu. Then there
exist w>0, M ^ l such that \\T(-t)\\^Mewt for all t^O. This implies that
\\T{t)f\\^M-'e-wt Il/H (feE). Hence for k><o{A), feE+,
\[e-»T(t)fdt =[e-"\\T(t)f\\dt
JO
\JQ
s*M- 1 \ e-Xte~wt
Il/H dt = ((A + w)M)~l \\f\\.
Jo
Similarly for /?(A, -A) where K>(o(-A).
from Theorem 2.5.
Thus (ii) holds. The converse follows
REMARK. Condition (2.5) does not hold for generators of positive Q-groups on
every Banach lattice [4, Example 2.2.13].
EXAMPLE 2.7. We show by an example that condition (2.5) cannot be omitted
in Theorem 2.6.
Let B be the generator of the group (T{tj)teU on LX(U) given by T{t)f(x) =
f(x + t). Then D(B) = {f eAC(U): f e L\U)} and
and
-*(-A, B)f(x) = R(k, -B)f(x) = e-** f
J—0
for A > 0, / e E. For n e N let
/>*(/) = (§)" I
\f(x)\dx
and EQ = {f e L\U): T>n=iPn(f) < °°}- Then EQ is a Banach lattice with the norm
H/llo := ll/lli + S"=iPn(/)- Of course, EQ is isomorphic to L\M, ju) for a suitable
measure JU. We show that /?(A, B)EQ <= EQ for all AeR\{0}. In fact, let A>0,
/ e EQ. Then
, B)f) ^ ($y fn
h-
e^ fe-» \f(y)\ dy dx
Jx
326
WOLFGANG ARENDT
Hence E~=ipn(#(A, Z?)/)<oo. Similarly for A<0. Let A be the operator on EQ
defined on D{A) = {/ e EQ D £>(£): Bf e EQ} by .4/ = Bf Then it is easy to see
that U\{0} c p(4) and fl(A, A) = R(k, B) \ EQ for A e R\{0}. Hence R(k, ^4) 5= 0
and /?(A, - A ) = -R{-k, A) s* 0 for A > 0.
We show that D(A) is dense in E. Let f EEQ. For n e N let /„ = / • 1R\[0>2-"]T h e n / - / n = / - l I 0 , 2 - ] . Hence \\f-fH\\x-+Q f o r n - ° o . Moreover, pm(f~fn) = 0
for m ^ n and /?„,(/ - / „ ) =pm{f) f° r w > n. Hence
for n -> oo. We have shown that EQQ : = {/ e EQ\ there exists e > 0 such that
/l[o,e] = 0} is dense in EQ. NOW let /e£oo- Then there exists e > 0 such that
/l[o e] = 0. It is easy to see that there exists a sequence (fn)czAC(U) such that
/; 6 L 1 ((R),/ n | [ o (e /2] = 0, and ||/-/ n || 1 - > Ofor/i->«>. Then/,, eD(A). Since || \\t
and || ||o are equivalent norms on Ee/2:= {feL1(R): f(x) = 0 for xe[O, e/2]
a.e.} CEQ, it follows that limn_»oo^ = / in EQ. Finally, we show that A is not a
generator. In fact, assume that there exists a semigroup (7o(f)),>o on EQ, which is
strongly continuous for t > 0, such that
R(k,A)f=re-x%(t)fdt
Jo
for large A > 0.
R(k,A)f=f e-KtT(t)fdt.
Jo
So it follows from the uniqueness theorem for Laplace transforms, that
T0(t)f = T(t)f (t > 0) for all / e EQQ. Let t > 1, n e N and / = 2nl[t+2-» t+2-n+l]. Then
||/||o=||/||i = 1. But 7o(0/ = 2"l[2-n,2-n+1]. Hence ||r o (0/llo^P.(r o (0/) = (1)"Thus 7o(0 is not continuous for t > 1.
3. Perturbation and examples
In this section we present two kinds of perturbations which demonstrate that
there exist many natural resolvent positive operators which are not generators of
a semigroup.
THEOREM 3.1. Let A be a resolvent positive operator and B\ D(A)-+E a
positive operator. If r(BR(k, A)) < 1 for some A > s(A), then A + B with domain
D(A) is a resolvent positive operator and s(A + B)<L
Moreover, if
sup{\\nR((i, A)\\: ^5=A}<oo (for example, if A is the generator of a Qsemigroup), then also sup{||///?(/*, A + B)\\: y, 2= A} <oo.
Note. By assumption, BR(k, A) is a positive, hence bounded operator on E;
we denote by r{BR{k, A)) its spectral radius.
Proof. Let / e D(A). Then
(A - (A + B))f = (/ - BR(k, A))(k - A)f.
RESOLVENT POSITIVE OPERATORS
327
Let Sk := (/ - BR(k, A))~l = E:=o (BR(kt A))n 2*0. Then
R(k,A)Sx(k-(A
+ B))f=f
forall/eD(>l)
and
(k-(A + B))R(k,A)SKg = g
forallge£.
Hence kep(A + B) and R(k, A + B) = R(X, A)Sk^0.
If ju>A, then
BR{n,A)^BR(k,A)
by (1.2), and so r(BR(n, A))^r(BR(k, A))<1- Hence
also pep(A + B) and R(fi,A + B)^Q. Moreover, S^^Sx and
R(X, A) so that
Hence
if the additional assumption is satisfied.
The following examples show that even in rather simple and natural cases
perturbations as in Theorem 3.1 may yield resolvent positive operators which are
not generators of semigroups.
EXAMPLE
3.2. Let a e (0,1). Define the operator A by
Af{x) = -f'(x) + (or/x)/(x)
(x 6 (0,1])
on the space E = Cb(O, 1] := {/e C[0, 1]: /(0) = 0} with domain D(A) =
{/e C^O, 1]: /'(0) =/(0) = 0}. Then A is resolvent positive but not a generator
of a semigroup. Moreover, s(A) = — oo and sup{||/xi?(ju, >l)||: jit ^=0} ^ 1/(1 — a).
Proof. Let Aof = -f with domain D(A0) = D(A). Then Ao is the generator of
the Cb-semigroup (T(t))tS:0 given by
f(x-t)
0
x>t,
otherwise.
Moreover, o(A0) = 0 and
R(k, A0)f(x) = e-** f ex>/(v) dy (AeC./e £).
•/o
Letfi: D(^ 0 )^^begivenby5/(x) = a/(jc)/x(^>0), B/(0) = 0. Let/e^and
g = R(0,A)f. Then
Jo
/(y) ^y
Thus ||BR(0, y40)|| ^ oc < 1. So Theorem 3.1 implies that A =A0 + B is resolvent
positive and s(A) <0. Moreover, for ju^O one has ju/?(/i, y4) = jui?(^, y40)5M ^
///?(/z,A0)50, where S^ = (/ - BR(n, Ao))-\ Since | | ^ ( ^ f i 4 o ) | | ^ l and ||5 0 || ^
1/(1 - a) it follows that sup{||//fl(ju, i4)||: \L ^ 0 } ^ 1/(1 - or).
It remains to show that A is not a generator. One can easily check that for all
328
WOLFGANG ARENDT
A e C one has A e p{A) and
fl(A, A)f(x) = e= {Xxa(x-tyy(x-t)e-ktdt
(feE).
Suppose that there exists a semigroup (T(t))t>0 which is strongly continuous for
t > 0 such that R{\, A)f = jo e~ktT{t)fdt for all/ e E and all sufficiently large real
A. Then by the uniqueness theorem for Laplace transforms, for 0 < t < 1, one
would have
r
•(x°l(x-t)")f(x-t)
»/w-l0
ioix^t,
otherwis.
This does not define a bounded operator on Q(0,1].
REMARK. It follows from a result of Benyamini [5] that Q(0,1] is isomorphic
as a Banach space to a space C(K) (K compact). Thus Example 3.2 yields an
operator B on C(K) such that a(B) = 0 and the resolvent satisfies
, B)"\\: X^O,
sup{||Afl(A, B)\\: A^
But B is not a generator. Of course, B is not resolvent positive by Corollary 2.3.
3.3. Let E = Lp[0, 1], where Kp <°°. Choose a € (0, (p - \)lp).
Define the operator A by
EXAMPLE
with domain D(A) = {feAC[Q, 1]: f'eLp[0, l],/(0) = 0}. Then A is resolvent
positive. Moreover, s(A)<0 and sup{||A7?(A, i4)||: A^=0}<<». But A is not a
generator.
Proof. Let Aof=—f
with domain D(A0) = D(A). Then Ao generates the
semigroup (ro(0),»o on E given by T0(t)f(x) =f(x -t)forx^t
and r o (O/(x) = 0
otherwise. Moreover, s(A0) = - » and /?(0, >lo)/(^) = 5of(y) dy. Let 5 :
E be defined by Bf(x) = (a/x)f(x). Then by [7, Lemma 1], BR(0, A0) e ( )
and ||Bi?(0, y4o)|| — aP/(p ~ !)• Hence Theorem 3.1 implies the first assertions
stated above. It remains to show that A is not a generator. It is not difficult to
check that the resolvent of A is given by
l eXyy-J(y)dy= \ xa(x - t)~J(x * dt (A>0).
o
h
Let Eo={feE: there exists 6 > 0 such that f(x) = 0 for almost all x e [0, 6)}.
Define T(t): EQ^E by T(t) = 0 if t s= 1 and
.0
otherwise,
if 0 s£ t < 1. Then T(-)/ is continuous from [0, °°) into E and
RESOLVENT POSITIVE OPERATORS
329
for all A 5= 0 if / e EQ. Thus if there exists a semigroup (Ji(t))t>0 which is strongly
continuous for t > 0 and such that
R{X,A)f=[e-*%(t)fdt (feE)
for large A, it follows from the uniqueness theorem for Laplace transforms that
Tx(i)f=T(t)f for all feEo, t^O. But for re(0,1) the mapping T(t) is not
continuous (from Eo with the induced norm into E). In fact, let )3 > 0 such that
1 - ocp < ftp < 1, and for neN, let /„(*) = l[1/n> x](x)x~p. Then /„ e EQ and
supfll/J,: neN}<cc.
But
=f
1
'fn(yy/yapdy
0
1-r
J\ln
>»
for n
since (a + fi)p > 1.
Next we consider multiplicative perturbations which will allow us to give
examples of resolvent positive operators on LF ( 1 ^ p<co) which are not
generators of semigroups.
3.4. Let (X, ju) be a o-finite measure space and E = LP(X, JJ,)
(1 ^p < °°) (respectively, X locally compact and E = Q ( i ) ) . Let A be a resolvent
positive operator. Suppose that m: X—>[0, <») is measurable (respectively,
continuous) such that m(x)>0 a.e. (respectively, m(x)>0 for all xeX) and
(l/m)feEforallfeD(A).
Let
PROPOSITION
D(A*) = {geE: m-geD(A),
(l/m)A(m
-g)eE)
n
and A g = (l/m)A(m • g). Then A* is a resolvent positive operator and s(An) ^
s(A).
Proof. For k>s(A) let R*(X)f= (l/m)R(k, A)(m •/) (feE). Then R*(k) is
a positive, and hence bounded operator. It is easy to show that /?#(A) =
3.5. Let E = Lp[0, 1], 1 =£/?<(», and A be given by Af-f
with
D(A) = {f eAC[0, 1]: f eLp[0, 1), / ( l ) = 0}. Then A is the generator of the
semigroup (T(t))t^0 given by
EXAMPLE
Moreover, s(A) = — <x and
k
\l^
(feE).
Let a e (0, 1/p) and m(x) = xa. Then 1/m e Lp[0, 1] and since D(A) c C[0, 1], it
follows that (l/m)/eZ/[0, 1] for a\\feD(A). By Proposition 3.4, the operator
330
WOLFGANG ARENDT
An is resolvent positive, where
D(A*) = {feE: m-feD{A), (l/m)A(m -f)eE]
and
A*f={\lm)A{m 'f) = x-a(xjy =/' + (afx)f.
The domain D{AU) is dense in L p [0,1]. In fact,
D{A) D {/ e Lp[0, 1]: /| [Oj e] = 0 for some e > 0} c: D(A*).
But D{A) is dense in Lp[0,1], and it is easy to see that every
= {geAC[0,l]: g'eZ/[0, l
can be approximated by functions in D(A) which vanish in a neighbourhood of 0.
We show that A** is not the generator of a semigroup. In fact, assume that
there exists a semigroup (T(t))t>0 which is strongly continuous for t>0 such that
for sufficiently large A. It is not difficult to see that
a kx
R{X, A*)f{x)=x- e
r1
Jx
ky
r1"*
a
e- f{y)y dy =
Jo
e-ktX-J{x + t)(x + t)°dt.
As in Example 3.3 one shows that for 0 < f < l , T(t) is given by T(t)f(x) =
x~a(x + t)af(x + t) for * + f=sl. This does not define a bounded operator on
Lp[0,1].
Note. For p > 1, R(X, A*) is the adjoint of R(k, A) in Example 3.3 on Lq(Q, 1)
(with Up + 11 q = 1). Thus XR(X, A*) is norm-bounded for A-»°°. This is also the
case for p = 1. In fact,
, A«)'\\ =
a
e-ky[yx-^dx\
y e [0,1]} = ||A/?(A)||,
where R(X) denotes the resolvent of the operator A on Qj(0, 1] in Example 3.2.
Hence ||A/?(A, A # )|| =s 1/(1 - a) (A 2*0).
REMARK. In the literature, the first example of a resolvent positive operator
which is not a generator was given by Batty and Davies [3] on Q)(1R). A similar
example on O(U) appears in [4, Example 2.2.11]. Independently, H. P. Lotz
constructed an example by a renorming procedure similar to Example 2.7
(unpublished).
4. Positive resolvent as Laplace-Stieltjes transform
In the sequel we continuously use the notion and properties of the LaplaceStieltjes integral as given in [15, Chapter III]. It is not difficult to see that any
increasing function S on [0, °°) with values in J?(£) (or E) is of bounded variation
[15, Definition 3.2.4]. Thus j^e~XtdS(t) is defined for every AeC, b^0. Then
the improper integral Joe~x'dS(t) is defined as Iim6_oo $oe~x'dS(t) in the
operator norm (cf. Proposition 5.5 below).
RESOLVENT POSITIVE OPERATORS
331
THEOREM 4.1. Let A be a densely defined resolvent positive operator. There
exists a unique strongly continuous family (S(t))tSsQ of positive operators satisfying
5(0) = 0 and S(s)^5(0 forO^s^t
such that
(4.1)
R(X,A)=\ e~ktdS{t) (k>s{A)).
The construction of Ax in the following proof is due to Chernoff [10].
Proof. Uniqueness of the representation (4.1) follows from [25,7.2]. In order
to prove the existence we first assume that s(A) < 0. Then for A > 0 one obtains,
from the resolvent equation,
(4.2)
R(0, A)kR(k, A) = R(0, A) -fl(A,A) ^ R(0, A).
Let ll/llj — inffllrtCOMtell: ±f^g}- Denote by Ex the completion of E with
respect to this norm. For / e E, A > 0 one has
\\kR(k, A)f\\x = inf{||*(0, A)g\\: ±kR{k,
^ifii{\\R(0,A)kR(k,A)h\\:
±f^h}
*inf{\\R{0,A)h\\: ±f^h} (by (4.2))
K.
Thus R(k, A) has a unique continuous extension RX(X) on Ex which satisfies
(4.3)
P*i(A)Hl
(A>0).
It is obvious that (/?i(A))A>0 is a pseudo-resolvent. Since D(A)a R1(X)E1 (A > 0),
it has a dense image, and so it is the resolvent of a densely defined operator Ax on
Ex [11, Theorem 2.6]. It follows from the Hille-Yosida theorem that Ax is the
generator of a strongly continuous contraction semigroup (7i(f))fs=o on Ex. The
operator R(0, A) satisfies
(4.4)
||K(0,i4)/||*||/||i
(feE).
(In fact, let/e£and ±f^g. Then ±R(0, A)f^R(0, A)g. Hence \\R(0, A)f\\ ^
\\R{0,A)g\\. Thus ||/?(0,i4]/||^inf{||/?(0,i4)g||: ±f^g} = \\f\\x.)
Consequently, the extensionfl^O)of R(0, A) onto Ex maps Ex into E. Moreover,
(fli(A))^o (A = 0 included) is a pseudo-resolvent too. Thus Rx(0) = R(0, Ax). This
implies that
(4.5)
D(i41) = ^(0,i4i)^ic£;.
The closure Ex+ of E+ is a cone in Ex which is invariant under /?(A, Ax) for A 5* 0.
This cone is proper (in fact, let / e Ex+ n ( - £ 1 + ) ; then R(0, Ax)f eE+C\ (-E+),
whence R(0, Ax)f = 0, and s o / = 0). Thus (EX, EX+) is an ordered Banach space
and the semigroup (Tx(t))tSl0 is positive.
Now let S(t)f = p0Tx(s)fdseD(Ax)^E
for feE. Then 5(0 is a positive
operator on E (t s* 0). It is clear from the definition that 0 = 5(0) ^ 5(5) «s 5(0 for
0<s<t. Moreover, let t>0. Then for feEx+,
332
WOLFGANG ARENDT
Hence
I
h
Thus,
(4.6)
= R(0, AJf-RiO, ^071(0/^^(0, A1)f.
S(t)*R(0,A) (t*0).
In particular, sup{||5(0||: f^0}<«>. We now show that 5 ( ) : [0, <*>)^><£(E) is
strongly continuous. Let f eD(A) and g = /!/. Then
\\S(t + h)f - 5(0/|| = ||*(0, A){S(t + h)g - S(t)g)\\
for h-* 0. Here we made use of (4.4). Thus 5(-) is strongly continuous on a dense
subspace. Since S(-) is bounded, this implies the strong continuity on the whole
space. Since 5(-) is bounded, the integral in (4.1) converges in the operator norm
for A > 0. Let / e E. Then
f e-k'dS(t)f= f e-kt
Jo
Jo
Thus (4.1) holds and the proof isfinishedin the case when s(A) < 0.
Now let s(A) be arbitrary. For w>s(A) consider the operator
B=A-w.
Then s(B)<0, so by what we have proved above, there exists a strongly
continuous increasing function Sw(-): [0, °°)-»i?(E) + satisfying 5^(0) = 0, such
that
for fj, > 0. Hence
J
.-00
/.00
I e- A 'e M "dS M ,(r)= e - A '
o
Jo
for all k>w, where 5(0 = Joe m dS w {s). Clearly, S(-) is strongly continuous,
increasing and satisfies 5(0) = 0. Because of the uniqueness theorem [25, 7.2], it
does not depend on w and so (4.1) holds for all A >s(A). This proves the theorem
in the general case.
EXAMPLE 4.5. (a) In order to illustrate the construction in the proof of
Theorem 4.1, consider the operator A* on Lx[0, 1] given in Example 3.5. Then
R(V,A*)f(x)=x-°[f{y)y°dy.
•Or
Thus
i=\\R{*,A*)\f\\\=\Xx-\\fiy)\y*dydx
Jo
Jx
= 1/(1 -a)C \f(y)\ydy.
Jo
RESOLVENT POSITIVE OPERATORS
333
Hence Ex = L\[0,1], (1 - a)^y dy) and
if JC ^ 1 - f,
otherwise,
0
for a\\feEu 12*0.
(b) Let A be the generator of a strongly continuous positive semigroup
( r ( 0 W T h e n S(')f = Jo T(s)fds for all/ e E, t& 0.
(c) If A is the operator in Example 3.2, then
I y~7(y)dy
ifx>t
Jx-t
5. Approach via Bernstein's theorem
A numerical-valued function is the Laplace-Stieltjes transform of an increasing
function if and only if it is completely monotonic (by Bernstein's theorem [25,
6.7]). Now let A be a resolvent positive operator. Then
()
, A)n+1 5*0
for all neN, k>s(A). Thus the function R(-,A) is automatically completely
monotonic whenever it is positive. And in fact we can use Bernstein's theorem to
prove the representation theorem (Theorem 4.1) if additional assumptions on the
space are made. On the other hand, it is not necessary to assume that A has a
dense domain.
DEFINITION
5.1. We say that E is an ideal in E" if for f e E, g e E", O^g =£/
implies geE.
Note. Here we identify E with a subspace of E" (via evaluation). Then
E+ D E = E+ (that is, £ is an ordered subspace of E").
LEMMA 5.2. Suppose that E is an ideal in E". Then the norm is order continuous
in the following sense. If (fn)ne^ is a decreasing sequence in E+, then
converges strongly (and limn_»oo/n = infneN /„).
Proof. Define FeE"+ by F(tf>) = infn6^</n, 0 ) (<t>eE'+). Then FeE" and
0^F^fn for all neN. Hence FeE by assumption. It follows from Dini's
theorem that limn^00</fl, 0 ) = (F, 0 ) uniformly on U%:= {<f> e E'+: | | 0 | | ^ 1 } .
Let # ( / ) : = sup{</, 0 ) : 0 e U%}. Since £ + is normal, \\f\\N: = N(f) +N(-f)
defines an equivalent norm on E (see [4]). But limn_>00||/n — F\\N = 0.
EXAMPLE 5.3. A Banach lattice E is an ideal in E" if and only if the norm is
order continuous, in the sense of Lemma 5.2 (see [22, II, §5]). For example,
LP(X, ju) ((X, fi) a (7-finite measure space and l=Sp<oo) and c0 have order
continuous norm, but C[0,1] has not.
334
WOLFGANG ARENDT
DEFINITION 5.4. A function/: (a, °°)-»E is called completely monotonic if / i s
infinitely differentiable and (-l)7 (/l) (A) ^ 0 for all A > a, n = 0,1, 2,....
Let or: [0, a>)->£V be an increasing function with values in E or
satisfying ar(0) = 0. We show that the Laplace-Stieltjes transform of or is a
completely monotonic function. Moreover, the abscissa of convergence does not
depend on which of the natural topologies is considered. In order to formulate
this more precisely, we say that the integral $oe~Xtdcc{t) converges weakly
(strongly, in the weak operator topology, or in the operator norm, respectively)
if Hindoo Jo e~kt d<x(i) exists in the corresponding topology.
5.5. Let w eU. Consider the following assertions:
(i) jo e~wt da(t) converges weakly;
(ii) for all X>w there exist M^0 and an element <*(<») of E {of ££{E),
respectively) such that \\ a(t) - or(oo)|| ^ Meh for all 15* 0;
(iii) |o e~M da(t) converges in the norm whenever Re A > w.
Then (i) implies (ii) and (ii) implies (iii). / / (i) holds, then the function
k-+ Jo e~Xt da(t) is completely monotonic on (w, °°). Moreover,
PROPOSITION
f e~Xtd(x{t)= \ ke~Xta{t)dt (ReA>max{0, w}).
h
Jo
Proof It follows by standard arguments that (ii) implies (iii) (cf. [15, Theorem
6.2.1]). We show that (i) implies (ii). It suffices to consider the case when a takes
values in E (the other case is treated analogously). At first we consider the case
when w s*0. Let A> w. Then for 0 €E+, t^0 one has
(5.1)
0^e-Xt{a(t), <t>)^e-kt(a(t), 0) + A I V ^ O ^ ) , <t>) ds
f
Jo
Xt
Hence sup,&0 e~ {a{t), <f>) <». Since E'+-E'+= E', it follows that (e~A'ar(f)),^0
is weakly bounded, and hence norm bounded. This proves (ii) in the case when
wssO. Now assume that w < 0 . Let p(t) = poe~wsa(s)ds. Then Jodj8(0 =
Jo e~ws da(s) exists weakly. Let 0 < d < -w. Then by the first case ||0(O|| ^ Me&
(f2=0) for some A/2*0. Moreover, ar(<»):= Jo doc(t) = Jo ews df$(s) exists (since
(iii) holds for p and A > 0). Hence
k(°°)-*(OII = M<
/•C
= -ew'P(t) + (-w)\
r
^(—w)M\ ewse ds
Jt
ewsP(s)ds\
RESOLVENT POSITIVE OPERATORS
335
Since 8 e (0, -H>) was arbitrary, this shows (ii) to hold. Furthermore, if (iii)
holds, then r(A) = [ e~kt doc(t) is C00 on (w, ») and
=f
(n = 0,1,2,...).
Finally (5.1) is proved by integrating by parts.
THEOREM 5.6. Assume that E is an ideal in E". Let f: (a, <*>)-* E be a
completely monotonic function. Then there exists a uniquely determined normalized increasing function a: (0, <*>)-* E such that
(5.2)
/(A) = f e~k'da(t)
Jo
REMARK. Here we call oc normalized if for all <f> e E'+ the function
h: t-+ (a(t), tf>) is normalized (that is, h(0) = 0, h{i) = \{h(t + 0) + h{t - 0 ) ) for
t>0).
Proof. Let (j> e E'+. Then A—• (/(A), <f>) is completely monotonic. So by
Bernstein's theorem there exists a unique normalized increasing function
c*4>: (0, «)-»R such that
(5.3)
From the uniqueness theorem [25, 7.2] it follows that oc^i) is additive and
positive homogeneous in (f> for every t^0. Thus for every f2=0 there exists
a unique oc^it) e(E')+ such that {$, ac(t)) = a^if) for all <peE'+. Let
A > max{a, 0}. Then for every 0 e E'+,
Consequently, a(t)^eXtf{X), and our assumption on E implies that a{t)eE+.
Since the integral in (5.3) converges for every k>a and <peE'+, we conclude
from Proposition 5.5 that the integral Jo e~Xt da(t) converges in the norm for
every k>a. Finally, (5.3) implies that
= fe- 'd(a(t),
x
Jo
<p) =([e-"d(x{t), cf>
\Jo
for all 0 e E'+. Hence (5.2) holds.
REMARK. There are other results related to Theorem 5.6. Schaefer [20]
obtained a characterization of completely monotonic sequences with values in an
ordered locally convex space as moments of an increasing function on [0,1]
(Hausdorff moment problem). Another vector-valued version of Bernstein's
theorem has been obtained by Bochner [6].
THEOREM 5.7. Suppose that E is an ideal in E". Let A be a resolvent positive
operator. Then there exists a unique strongly continuous family (S(t))tSt0 of
336
WOLFGANG ARENDT
operators on E such that
0 = 5(0)^5(5)^5(0
and
R(k, A) = f e~k'dS(t) (A>j(i4)).
h
Proof. Uniqueness follows from [25, 7.2]. We show the existence of the
representation (5.4). Let feE+. Then R(-,A)f is a completely monotonic
function from (s(A), <») into E. By Theorem 5.6 there exists a unique normalized
increasing function 5(-,/): (0, <»)-»£ such that
(5.4)
R(k,A)f=\
(5.5)
e-x'dS(t,f)
Jo
From the uniqueness theorem it follows that for every t^O the mapping
f—*S(t,f)
from E+ into E+ is additive and positive homogeneous. Since
E = E+ — E+, there exists a unique linear operator 5(0 on E such that
S(t)f = S(t,f) for all / e £ + . Since 5(-)/ is increasing for all feE+, 5(-) is
increasing. Let fi>s(A). Then it follows immediately from the definition that
R((i, A)S(-)f and 5(-)/?(/u, A)f are also normalized for all f e E+. Moreover, for
allA>s(,4),
f e~k'd{R{n, A)S{t)) = R{fji, A)R(k, A)
Jo
e-ktd(S(t)R(k,A)).
= R(k,A)R(fi,A)=
Jo
Hence it follows from the uniqueness theorem that
(5.6)
S(t)R(ii, A) = R(n, A)S(t)
(t^O).
Now let / e D{A). Then for all A > mdx{s(A), 0},
f X2e-Xltfdt
Jo
= I k2e~XtS(t)fdt- I ke-x'S(t)Afdt
Jo
Jo
= f k2e~ktS(t)fdtJo
[
Jo
k2e-kt\s{s)Afdsdt.
JO
Thus
[ e~kt(tf - 5(0/ + [ S(s)Afds) dt = O
for all A>max{0, s(A)}. Consequently, by the uniqueness theorem,
(5.7)
S(t)f = tf+\'s(s)Afds
(**0).
Jo
This implies that 5(-)/ is continuous for all / e D(A). Now let g e E+, t>0. Then
lini^, S(s)g = :h+ and lim^, S(s)g = :h_ exist by Lemma 5.2. We have to show
RESOLVENT POSITIVE OPERATORS
337
that h+ = h.. Let X>s(A). Then by (5.6),
R(k, A)h+ = lim 4 , R(k, A)S(s)g = lim 4 , S(s)R(X, A)g
= S(t)R(k, A)g
(since R(X, A)g e D(A))
= lim,tl S(s)R(k, A)g = R(k, A)(\im S(s)g)
=
R(k,A)h-
Since R(X, A) is injective, it follows that h+ = h-.
6. The integrated semigroup
Let A be a resolvent positive operator. We assume that there exists a strongly
continuous increasing function 5: [0, <»)—» J£(£) satisfying 5(0) = 0 such that
R(k, A) = I e~Xt dS(t)
(6.1)
(A > Ao)
for some ko^s(A).
By the results of the two preceding sections such a
representation of R(X, A) exists when either A is densely defined or E is an ideal
in E". Note that (S(t)),^0 is uniquely determined by A, and we call (S(O)rs=otne
integrated semigroup generated by A. (Of course, this terminology is motivated by
the case when A is the generator of a strongly continuous semigroup (T(t))t9,0
because then S(t) = Jo T(s) ds.)
The following proposition shows that s(A) is determined by the asymptotic
behaviour of S(t) for f-»a>.
PROPOSITION
6.1. (a) If keC
R(X,A)=
\
satisfies Re A >s(A),
then Xep(A)
and
e~XtdS(t)
Jo
(6.2)
=
Xe~XtS{t) dt if in addition Re A > 0.
Moreover, s(A) e o(A) (the spectrum of A) whenever s(A) >°°.
(b) If s(A)^0, then s(A) = inf{w > 0 : there exists Af^O such that \\S(t)\\^
Mewlforallt^0}.
(c) // s(A) < 0, then lim,^ S(t) = R(0, A) and s(A) = inf{w> < 0: there exists
M^Osuch that \\R(0, A) - S(t)\\^Mewt for allt^O}.
Proof For feE+, <j> e E'+ denote by sf<p the abscissa of convergence of the
integral $%e~Xtd{S(i)f, $). Then by [25, Chapter 5, Theorem 10.1], sf4> is a
singular point of the analytic function A-* Jo e~ktd(S(t)f, <f>) (ReA>s).
Consequently,
s:=sup{sf4): f eE+, (f> eE'+}^s(A).
Since
||/?(A, A)\\ 5=
dist (A, a ^ ) ) " 1 for all A e o(A), it follows from the uniform boundedness
principle that A e p(A) whenever Re A >s(A). So (6.2) holds by Proposition 5.5.
Suppose that -°° < s(A) $ o(A). Then (s(A) - 6, oo) c p(A) for some 6 > 0 and
so s^s(A)-6.
Hence (R{k, A)f, <p) 2*0 for all feE+, <f> eE'+, k>s(A) - d.
This implies that R(A, A) 2= 0 for A > s(A) - 6, which contradicts the definition of
338
WOLFGANG ARENDT
s(A). This finishes the proof of (a). Assertions (b) and (c) follow directly from
Proposition 5.5.
REMARK 6.2. (a) It follows from Proposition 6.1 that s(A) is the abscissa of
convergence of the Laplace-Stieltjes transform (6.2) (no matter whether the
weak or strong operator topology or the operator norm is considered, see
Proposition 5.5).
(b) If A is the generator of a positive Co-semigroup, then Proposition 6.1
implies that R(k, A) = Jo e~ktT{i) dt, where the integral converges in the operator
norm, for all AeC such that ReA>s(,4). (Here lbQe~XtT{t)dt is defined
strongly.) However, it may happen that s{A) < a>(A) (see [14]). Thus s(A) is
determined by the asymptotic behaviour of the integrated semigroup but not of
the semigroup.
Now we establish the relations between A and the integrated semigroup. The
operators A and S(t) commute. In fact,
(6.4)
S(t)R(k,A) = R(k,A)S(t)
(k>s(A), t>0).
(This is proved as (5.6).) As a consequence,
(6.5)
feD(A)
implies S(t)feD(A) and AS(t)f = S(t)Af (t^O).
Given / e E, by a solution of the inhomogeneous Cauchy problem
v'(t)=Av(t)+f (t^O),
v(0) = 0,
we understand a function v e C^QO, °o), E) satisfying v(t) e D(A) for all f ^ 0 such
that (6.6) holds.
PROPOSITION 6.3. Let f e D{A). Then v{i) = S(t)f (t s* 0) defines a solution v of
(6.6). Conversely, if feE and v is a solution of (6.6), then v(t) = S(t)f for all
t^O.
Proof The proof of Theorem 5.7 shows that (5.7) holds, and consequently
S(-)f is a solution for every f e D(A).
Conversely, let f e E and assume that v is a solution of (6.6). Let t > 0. Define
w(s) = S(t - s)v(s) (s e [0, t]). Since v(s) e D(A) by hypothesis, it follows that
w'(s) = -AS(t - s)v(s) - v(s) + S(t - s)v'(s) = -v(s) + S(t - s)f.
Consequently,
0 = w(t) - w(0) = f (S(t - s)f - v(s)) ds.
Jo
Hence Jo v(s) ds = S'o S(s)fds (t ^ 0). This implies that v = S(-)f
REMARK. The inhomogeneous Cauchy problem (6.6) has a unique solution for
all / e E if and only if A is the generator of a Co-semigroup (see [9, Theorem
3.1]).
PROPOSITION 6.4. For all f e E, f s* 0 one has
\s(s)fdseD(A)
h
and A\S(s)fds = S(t)f-tf.
Jo
RESOLVENT POSITIVE OPERATORS
339
Proof. Let A e p(A) and
v(t) = A f
tR(k, A)f - S(t)R(k, A)f (t ^ 0).
Jo
Then v(t) e D{A) (by (6.4)) for all t ^ 0. Moreover, u(0) = 0 and it follows from
Proposition 6.3 that v'(t) = S(t)f (t ^0). Consequently, JoS{s)f ds = v(t) e D(A)
and
,
,
(A-,4) S(s)fds = (X-A)v{t) = X\ S(s)fds + tf-S(t)f
Hence A /{, S(s)fds = S(t)f - tf.
PROPOSITION
(6.7)
6.5. Lets, t>0. Then
5(5)5(0 = f S(r) dr - fs(r) dr - f S(r) dr.
Jo
Jo
h
In particular, S(s)S(t)f e D(A) for all f e E and
AS(s)S(t)f = S(s +
t)f-S(s)f-S(t)f.
Moreover, S(s)S(t) = S(t)S(s).
Proof. Let ^ > 0, / e E. For 15* 0 let
t/(0= T V ) / , * / - - fs(r)fdr- fs(r)fdr.
Jo
Jo
Jo
Then by Proposition 6.4, v(t)e D(A) and
Av(t) = 5(5 + 0 / - 5(5)/ - 5 ( 0 / = v'(0 - 5(5)/
Since u(0) = 0, it follows from Proposition 6.3 that v(t) = S(t)S(s)f (t 5= 0).
The functional equation (6.7) corresponds to the semigroup property. In fact,
assume that (r(0) t >o is a strongly continuous family of operators such that
5(0:= Jo T(r)dr exists strongly. Then (5(0)^0 satisfies (6.7) if and only if
T(s + t)= T(s)T(t) for all s,t>0. This leads us to the following definition.
DEFINITION 6.6. A strongly continuous family (5(0)r^o in !£(E) is called
integrated semigroup if 5(0) = 0 and (6.7) holds for all 5, t > 0. Moreover,
(5(0)^0 ls ca^e<^ non-degenerate if for all 0i=feE there exists t > 0 such that
5(0/*0.
Note that the integrated semigroup generated by a resolvent positive operator
is non-degenerate and increasing. Next we show that conversely, every nondegenerate, increasing integrated semigroup is generated by a resolvent positive
operator.
The following result is due to H. Engler (oral communication) after a special
case had been proved by the author.
6.7. Let (5(0)^0 °e °n increasing integrated semigroup. Then there
exist M^0 and weU such that \\S(t)\\ ^Me wt for all t^0.
PROPOSITION
340
WOLFGANG ARENDT
Proof. For all t ^ 0 we have
f
S(r) dr = \ S(r) dr + f S(r) dr + 5(2)5(0-
Jo
Jo
Jo
Hence
rl+2
S(t +1) =s f
S(r) dr = [ S(r) dr + 5(2)5(0-
Jt
Jo
Consequently, there exists q>\ such that \\S(t +1)|| ^q(l + ||5(0
Let w: =
Hence, ||5(n)|| ^q + q2 + ... +qn^Nqn (neN) for N: = q/(q-l).
l o g q a n d M :=kNew
( w h e r e k is g i v e n b y ( 1 . 2 ) ) . T h e n f o r t ^ 0 t h e r e exist neN
and ^ e [0,1) such that t = n + s. Thus
||5(0N*||5(n + l)||
6.8. Let (5(0)rs=o ^ e fl non-degenerate, increasing integrated semigroup. Then there exists a unique resolvent positive operator A such that
R(k, A) = Jo e~kt dS(t) (A > s{A)). Moreover, A is given by D(A) = {feE: there
exists a (necessarily unique) geE such that 5(0/ = tf + Jo S(r)g dr for all t s* 0}
and Af = g.
THEOREM
Proof. By Proposition 6.7 there exist Af^O, w^O such that
(t^0). Let
J
i.00
/.00
e~k'dS{t)= \ ke~XtS(t)dt
Jo c p(A) and i?(A, >!) = /?(A) for all
Define A as above. Weoshow that (w,»)
A > w. L e t / 6 D(A). Then (d/dt)S(t)f = f + S(t)Af (t ^ 0); hence
- . 4 ) / = f k2e~ktS(t)fdtJo
f Ae"A'
Jo
r00
W
r00
"^
Jo
J Ae- '-5(0/<frx
o
= f ke-kl(S(t)Af+f)dtJo
te-u
I ke~ktS{t)Afdt=f
(A>w).
Jo
Conversely, let A > w, h e E. We claim that R(X)h e D(A) and (A - A)R(X)h = h.
For f 2s 0 we have
15(0* (A)/i = J V ^ | (5(05(5)^) ds
= I Ae";u(5(f + s) - S(t))h ds (by (6.7))
Jo
= f ke-^Sit + s^ds- S(t)h
Jo
= f ke'^iSit + s)- S(s))h ds + R(k)h - S(t)h
Jo
RESOLVENT POSITIVE OPERATORS
= Jof
= JoJoI
341
r) - S(r))h drds + R{X)h - S(t)h
/* ds + R(k)h - S(t)h (by (6.7))
= S(t){kR{X)h-h)
By the definition of A this implies that R{X)h e D(A) and AR(k)h = A/?(A)/i - h.
We conclude this section by studying differentiability of the integrated
semigroup in 0. Let (S(O)<&o be a non-degenerate increasing integrated semigroup with generator A. It follows from Proposition 6.3 that (d/dt)\t=0S(t)f=f
for a l l / e D(A); that is,
(6-8)
lim(l/0S(0/=/
PROPOSITION
(fzDW)-
6.9. The following assertions are equivalent:
(i) supo^i r 15(011 <»;
(ii) lim sup^co ||Ai?(A, i4)|| < «.
Moreover, (i) /io/ds anrf D(y4) is dense if and only if
(6.9)
lim-5(0/=/ forallfeE.
tio t
Proof. Let w e R. Then Condition (i) as well as Condition (ii) holds for A if
and only if it holds for A — w (observe that the integrated semigroup (Sw(t)),^0
generated by A — w is given by
Sw(t) = f e~ws dS(s) = e~wtS(t) + w f e~swS(s) ds).
Thus we can assume that s(A) < 0.
Assume that (i) holds. Then M:= sup0<,<oo(l/0 11^(011 <O °- Hence
[ k2te~x'dt =
for all A>0. Conversely, assume that supAss0 ||Ai?(A, ^4)|| <«». Let t>0. Choose
A = r"1. Then
is(0 = - (dS(
t
t Jo
This implies that supf&0 f"111^(011 < °°- Moreover, by Proposition 6.5, S(t)S(s)f €
D(A) for a l l / e E and \imti0 r1S(t)S(s)f = S(s)f Hence (6.9) implies that D(A)
is dense. The converse follows from (6.8).
REMARK. The argument in the proof is due to G. Greiner (in the context of
positive semigroups, cf. [17, C-III, Definition 2.8]).
EXAMPLE 6.10. (a) The operators given in Examples 3.2, 3.3, and 3.5 satisfy
the equivalent conditions of Proposition 6.9. Moreover, they all are densely
defined.
342
WOLFGANG ARENDT
(b) Consider the operator -A, where A is denned as in Example 2.7 on
Let (S(O)rso be the integrated semigroup generated by —A. Then
EQCIL1^).
S(t)f(x)=ff(x-s)ds=\X f{y)dy.
JO
1
We show that lim,.^'" ||S(0ll
Jx-t
=0
Let
°-
^
= 2
~"
and
/„ = 2 n l I _ 2 - 0 ]- Then
ll/-llo= ll/nlli = I- Moreover,
1
0
for x *£ -2~n,
2nx 4- 1
^ xT l
for — 2~n < x ^ 0
lor L
^x^\),
X
1 U 1 \J ^*^ Ar ***** &
4* Ar
0
Hence
y
for2-"<x.
n+l f2""
J2_n_i
Hence,
f-) = ^) n + 1 -*«>
forn^oo.
7. 77ie homogeneous Cauchy problem
Let >1 be an operator on E and / e D(A). By a solution of the homogeneous
Cauchy problem
(71)
u'(t)=Au(t) (^0),
"(0)=/,
we understand a function w e C^fO, °°), £) satisfying w(f) e D{A) for alU ^ 0 such
that (7.1) holds.
THEOREM 7.1. Assume that A is resolvent positive and either D(A) is dense or E
is an ideal in E" (see § 5). For every f e D(A2) there exists a unique solution of the
Cauchy problem (7.1). Furthermore, denote by (S(0)»3=o the integrated semigroup
generated by A. Then u(t) = S(t)Af + f for all t^O. Moreover, if /2*0, then
u(t) ^ 0 for all 12s 0.
The solutions of (7.1) depend continuously on the initial values in the following
sense: let fn e D(A2) such that lim,,.^/, = / in the graph norm. Denote by un the
solution of (7.1) for the initial value fn. Then un{t) converges to u(t) in the norm
uniformly on bounded intervals.
Proof. Uniqueness follows from Proposition 6.3. In order to prove existence
we assume that s(A) < 0 (otherwise one considers A - w instead of A for some
w>s(A)). Denote by (S(t))tSe0 the integrated semigroup generated by A. Let
feD{A2) and define u{t) = S(t)Af+f (t ^ 0 ) . Then by Proposition 6.3, u'(t) =
AS{t)Af + Af = Au{t) (f 2=0). Thus u is the solution of (7.1). Now let/„ eD(A2)
such that lmv^oo/n = / i n the graph norm. Let un(t) = S(t)Afn +fn. Since (S(t))tst0
is strongly continuous, it follows that un(t) converges in the norm to u(i)
uniformly on bounded intervals. Finally, assume that 0^f eD(A). Then using
RESOLVENT POSITIVE OPERATORS
343
(6.5) and Proposition 6.3 one obtains
u(t) = S(t)Af +f = AS(t)f +f = j t (5(0/).
Hence u(t) 5=0, since 5(-) is increasing.
REMARKS
7.2. (a) If D(A) is dense, then D(A2) is also dense. In fact, let
Aep(/l); then
E = D(Ay = (R(X, A)E)~ = (R(k, A)D{A)-y c (K(A, A)D(A))~
(b) In general, there does not exist a continuously differentiable solution of
(7.1) for every initial value in D(A). In fact, if D(A) is dense, this would imply
that A is the generator of a Q-semigroup (see [13, II, Theorem 1.2]).
(c) The continuous dependence of the solutions on the initial values is no
longer guaranteed if in Theorem 7.1 one replaces the graph norm by the norm. In
fact, if D(A) is dense, this implies that for every t =s 0, the operator T0(t) given by
T0(t)f = S(t)Af+f (from D(A2) into E) has a continuous extension T(t) on E. It
is not difficult to see that then (T(t))t3sQ is a C^-semigroup whose generator is A.
(d) Under the hypothesis of Theorem 7.1 the number s(A) determines the
asymptotic behaviour of the solutions of (7.1). In fact, suppose that s(A)<0.
Then by Proposition 6.1, \\S(t) - R(Q, A)\\ ^ Me'* (t ^ 0) for some 6 > 0. Hence
IK0H = \\S(t)Af+f\\ = 115(0^-^(0, A)Af\\ ^Me-* \\Af\\ (t^0).
Thus the solutions tend to 0 for f-»oo. (See [17, C-IV, §1] for this and related
results in the context of positive Q-semigroups.)
Next we prove a converse of Theorem 7.1.
THEOREM 7.3. Let A be densely defined such that p(A)=£0. Assume that the
following conditions hold:
(a) for every f e D(A2) the Cauchy problem (7.1) has a unique solution u, and
u{t)250 for all t^0 whenever f ^ 0 ;
(b) D(A2) C\E+is dense in E+.
Then A is resolvent positive.
REMARKS. 1. By Theorem 7.1 Condition (a) is also necessary. We do not know
whether every densely denned resolvent positive operator A satisfies (b). It is
certainly the case if in addition lim sup^*, ||A/?(A, A)\\ < <».
2. If A is a resolvent operator such that D(A)+ is dense in E+, then (b) holds
as well since limA_*oo A/?(A, A)f =/for all / e D(A).
We first prove a lemma.
LEMMA 7.4. Let A be an operator satisfying p(A)^0.
The following assertions
are equivalent:
(i) for every f e D(A2) there exists a unique solution of the homogeneous
problem (7.1);
344
WOLFGANG ARENDT
(ii) for every g e D(A) there exists a unique v e (^([O, o°), E) satisfying v(0) = 0
and v{t) e D(A) for allt^O such that v'(t)= Av(t)
+g(t^0).
Proof Let A e C. If u is as in (i), then u1(t) = e~Xtu(t) defines a function
satisfying u[(t) = (A - fyu^t), w^O) =/. If v is a function as in (ii), then
= e~Ktv(i) + A f e-^vis) ds
defines a function satisfying v'1(t) = (A - k)vx(t) + g, vl(0) = g. Thus A satisfies
(i) or (ii), respectively, if and only if A - A satisfies (i) or (ii), respectively. Since
we can assume that 0ep(A) (considering A-k if necessary). Let
p(A)^0,
g e D(A) and u e (^([O, «>), E) such that u{t) e D(A) (t ^ 0). Let v(t) = u(t) -
A~xg. Then u'(t)-Au(t)
(t^O) and u(0)=A~1g if and only if v'(t)-Av(t)
+g
(t 25 0) and v(0) = 0. This proves the claim.
Proof of Theorem 7.3. By Lemma 7.4 the assumptions of the theorem imply
there exists a unique solution v(-,f) of the inthat for every feD(A)
homogeneous problem (6.6). Let E1 denote the Banach space D(A) endowed
with the graph norm || H^. For feE1 let Si(t)f = v(t, / ) ; then the uniqueness of
the solutions of (6.6) implies that Si(t): E^E-^ is linear for every f2*0.
Moreover the mapping Si(*)/from [0, °°) into Ex is continuous for every feEx.
We show that Si(t) is a bounded operator for every t ^ 0.
Consider the Fr6chet space C([0, °°), Ej) of all continuous /^-valued functions
on [0, °°) topologized by the family (pn)neN given bypn(f) = supo«f<n ||/(0lU- Let
W: Ex-+ C([0, oo), Ex) be defined by Wf = Sx(-)f. Then Wis linear. We show that
W is closed. Then it follows that W and so also S^f) is continuous (t ^=0).
Let limn_»oo/n = / i n Ex and suppose that limn_^oo Wfn = v in C([0, <»), Ex), Then
limrt_oo Sx(t)fn = v(t) and hmn^^ASxit)^ =Av(t) in the norm of E uniformly on
compact intervals. Since
it follows that v(t) = tf + $'0Av(s) ds for all t^O.
From the uniqueness of the
solutions of (6.6) (by Lemma 7.4) it follows that v(t) = v(t,f) = Sx(t)f Hence
v = Wf
Next we show that (5'1(f)),S!0 satisfies the functional equation (6.7). Let
feElts>0.
For f>01et
w(t) = f Sx{r)fdr - \ Sx(r)fdr - \ Sx(r)fdr.
Jo
Jo
Jo
Then w(0) = 0 and
w'(t) = Sx(t + s)f - Sx(t)f = Aw(t) + Sx(s)f.
Hence w(t) = Sx(t)Sx(s)f (t^0) by the uniqueness of the solutions of (6.6). We
have shown that (Sx(t))tSi0 is an integrated semigroup on Ex. It follows from the
definition that it is non-degenerate. Now we define the integrated semigroup
{S(t))t^o on E by the similarity transformation S(t) = (ju - A)S1(t)R(fi, A) where
ju e p(A) is fixed (observe that (jU —A) is an isomorphism from Ex onto E).
RESOLVENT POSITIVE OPERATORS
We claim that 5(0/ = ^ ( 0 / for feD(A),
, A)v(t, /). Then
f2*0. Let feD(A)
345
and w(0 =
w(0) = 0, w'(t) = R(n, A)Av(t, f) + R(n, A)f = Aw{t) + R(ji, A)f.
Hence w(t) = v(t, R(n, A)f); that is,
5(0/ = 0* - A)v(t, R(n, A)f) = v(t, / ) (/ ^ 0).
This proves the claim. As a consequence, R((i, A)S(t)f = S(t)R((i, A)f whenever
/ e E. This in turn implies that 5(0/ € D{A) and .45(0/ = S(t)Af ior a11 / e ^ ( ^ ) We show that (S(t))tss0 is increasing. Let / e D(A2) fl £+. By hypothesis there
exists a positive solution u of (7.1). Let u(f) = Jo w(s) iy. Then v(0) = 0 and
= (
Thus u(0 = 5(0/ (^^0). Consequently S(s)f = v(s)^v(t) = S(t)f whenever
0=ej^^, feD(A2)C\E+.
Since D(A2)HE+ is dense in £"+, this implies that
(5(0)«*o is increasing. Denote by B the generator of (5(O)r;*o- It remains to show
t h a t B = A UfeD(A), then
Hence feD(B)
Then
and Bf-Af
by the definition of 5. Conversely, let/eD(J3).
Hence
, A)Bf + R(>i, A)f =
Since (S(t))CSsQ is non-degenerate, this implies that R(n, A)Bf - AR(n, A)f =
ju/?(/i, A)f -f. Hence / e D(A) and Bf = >4/.
8. The inhomogeneous Cauchy problem
Let A be a resolvent positive operator. We assume that A has a dense domain
or that E is an ideal in E" (see §5). Let F: [0, <»)—>£ be a continuous function.
Given / e E, we consider the inhomogeneous Cauchy problem
u'(0 = Au(0 + F(0
u(0)=/.
(f>0),
By a solution of (8.1) we understand a function u e £^([0, °°), £") satisfying
u(t) e D(A) for all t^ 0 such that (8.1) holds.
8.1. Assume that F e C2([0, «), £). / / / G D(A) such that Af + F(0) e
D{A), then the inhomogeneous problem (8.1) has a unique solution.
THEOREM
346
WOLFGANG ARENDT
Proof. Denote by (S(t))tSi0 the integrated semigroup generated by A. Asssume
that u is a solution of (8.1). For s e [0, r] let v(s) = S(t ~s)u(s). Then by
Proposition 6.3,
v'(s) = -AS(t - s)u(s) - u(s) + S(t - s)u'(s) = -u(s) + S(t - s)F(s).
It follows that
- 5 ( 0 / = v(t) - v(0) = - ( u(s) ds + I S(t - s)F(s) ds.
•'0
TT
•'0
Hence
f u(s) ds = S(t)f + f S(s)F(t - s) ds.
Jo
h
Differentiating and using Proposition 6.3 one obtains
u(t) = 5(0(Af + F(0)) + / + fs(s)F'(t -s)ds (t* 0).
h
This shows uniqueness. In order to prove existence define u by (8.2). Then
u e C\[0, oo), E) and
(8.2)
M'(0 = AS(t)(Af + F(0)) + Af + F(0) + S(t)F'(0) + f S(s)F"(t - s) ds.
Jo
Let us show that u{t)eD{A).
Moreover,
We have S(t)(Af + F(O))+feD(A)
f S{s)F'{t -s)ds=(
Jo
by (6.5).
S{s)F'(0) ds t f S{s) \ V"(r) dr ds
Jo
Jo
Jo
= f S(s)F'(0) ds+( ( S(s)F"(r) ds dr.
h
h Jo
It follows from Proposition 6.4 that this term is in D(A) and
A I S(s)F'(t -s)ds = S(t)F'(0) - tF'(Q) + [ (S(t - r)F"(r) -(t- r)F"{r)) dr
Jo
Jo
= 5(0F'(0) - fF'(O) + fs(r)F"(t - r) dr + fF'(O) - [F'{r) dr
Jo
Jo
= 5(0F'(0) + f S(r)F"(t -r)dr-
F{t) + F(0).
Hence u(t)eD(A) and
Au(t) = AS(t)(Af + F(0)) + Af + 5(0F'(0) + f S(r)F"(t -r)dr-
F{t) + F(0)
Jo
= u'(t)-F(t).
Hence (8.1) holds.
9. A characterization of resolvent positive
operators by Kato's inequality
Up to this point we assumed that a resolvent positive operator was given. Now
we find conditions on A which imply that A is resolvent positive.
RESOLVENT POSITIVE OPERATORS
347
Throughout this section we assume that £ is a Banach lattice with order
continuous norm and that there exists a strictly positive linear form </> on E. Then
II/IU :== (I/I' 0 ) d e f i n e s a n o r m o n E' W e denote by (E, <p) the completion of E
with respect to this norm. Then (E, <(>) is an AL-spacc (and is isomorphic to a
space of type L 1 [22, II.8.5]). Moreover, E is an ideal in (E, 0); that is, if
f,ge(E, <f>), \g\ =£/, a n d / e E , then also geE (see [22, IV.9.3]). For example,
let E = LP(X, y.) (l«sp<°°), where (X, JU) is a a-finite measure space. Let
<t>eLq(X,u) (where (1/p) + (1/?) = 1) and 4>(x)>0 fi-a.e. Then (E,<j>) =
L\X, 0/z).
F o r / e E we denote by sign/the unique operator on E satisfying (sign/)/ =
|/|, |(sign/)g| *£ |;| (g e E), (signf)g = 0 if |/| A \g\ = 0 (see [17, C-I, §8]).
THEOREM 9.1. A densely defined operator A on E is resolvent positive if and
only if the following assertions hold:
(K) there exist a strictly positive 0 e D(A') and koeU such that A'$ «s AO0 and
((signf)Af, <t>) *s <|/|, A'<f>) (feD(A))
(Kato's inequality);
(R) (ju0 — A)D(A) = E for some jU0>A0 (range condition). Moreover, in that
case, A is closable in (E, <f>) and its closure is the generator of a positive
CQ-semigroup on (E, (p).
REMARKS, (a) Condition (K) involves an abstract version of Kato's inequality.
See [17, Chapter A-II] and [1] for further information and the relation to the
classical inequality.
(b) In some aspects the theorem is similar to the Lumer-Phillips theorem [11,
Theorem 2.24]. The condition that A is dissipative is replaced by Kato's
inequality and the existence of a strictly positive subeigenvector oiA'. In contrast
to dissipativity, this condition is satisfied by A if and only if it holds for A + w
(weU).
Proof Assume that the conditions of the theorem hold. Considering (A — Ao)
instead of A we can assume that Ao = 0. Let N(f)= (f+, (f>) for all feE. The
function N is the restriction to E of the canonical half-norm (see [2]) on (E, <p).
By [1, Proposition 2.4] it follows from Kato's inequality that A is N-dissipative
[2]. Since D(A) is dense in E, it is also dense in (E, <p). Thus it follows from [2,
Theorem 2.4] that A is closable in (E, (j>) and the closure Ax oiA is 7V-dissipative.
Since E = (tio-A)D(A)ci([Ao-A1)D(Al),
HQ-AX
also has dense range. So it
follows from [2, Remark 4.2] (see also [19]) that Ax generates an N-contraction
semigroup, that is, a positive contraction semigroup on (E, <p). In particular, Ax
is resolvent positive and ^(-A^^O. It follows from (R) that noep(A) and
R(fi0, A) = R((x0, A^E. Moreover,
(9.1)
Af = AJ
s
and D(A) = {f e D(Al)C\E:
AJeE}.
Let n 3 jU0- Then for f e E+, R(fi, Ax)f *£ /?(ju0, At)f e E. Since E is an ideal in
(E, <p), it follows that R(n, Ax)EczE for all JU^/IQ. This together with (9.1)
implies that ft e p(A) and R(n, A) = R(fi, AX)\E for all ju^// 0 - Thus A is
resolvent positive and s(A) =£ /z0.
Conversely, let A be resolvent positive. We may assume that s(A) < 0. Then it
can be seen from the proof of [1, Proposition 1.5] that there exists a strictly
positive <peD(A') such that A'<£*£(). Consider the canonical half-norm N on
348
WOLFGANG ARENDT
(E, <t>). Then
N(kR(k, A)f) = ((A/?(A, A)f)+,
0 ) ^ <Ai?(A, A)f+, 0 )
for all / e E, A > 0. Hence i4 is N-dissipative. Since (1 - A)D(A) = E is dense in
(£, 0), it follows from [2, Remark 4.2] that >l is closable in (E, 0) and its closure
generates a positive semigroup on (E, 0). Hence (K) follows from [1, Proposition
1.1] (observe that f o r / , g e E , by the uniqueness of the signum operator the
vector (sign/)g is the same no matter whether (sign/) is considered as an
operator on E or (E, 0)).
REMARK 9.2. If A is a densely defined resolvent positive operator such that
lim supA_oo ||A/?(A, A)\\ <°°, then
holds for all / eD(A), xp eD(A')+. This can be proved in the same way as [1,
Proposition 1.1] if T(t) is replaced by (1 - tA)~x (t>0 small), because
(1-L4)" 1 / for all
=o
feD(A).
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